Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int N = 1e5 + 15;
const int maxn = 2e5 + 14;
const int INF = (1 << 30) - 1;
const int letter = 130;
const double pi = acos(-1.0);
const double eps = 1e-8;
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, m, q, tot, head[N];
int tree_idx, size[N], top[N], son[N], pre[N], deep[N], pos[N];
struct edges {
  int to, next;
} e[N << 1];
struct node {
  int num;
  int ps[13];
  node() {
    num = 0;
    memset(ps, 0, sizeof(ps));
  }
} dp[N], ans[N << 2];
void addedges(int u, int v) {
  e[tot].to = v, e[tot].next = head[u], head[u] = tot++;
}
void init() {
  tot = tree_idx = 0;
  memset(head, -1, sizeof(head));
}
void dfs1(int x) {
  son[x] = 0, size[x] = 1;
  for (int i = head[x]; i != -1; i = e[i].next) {
    int to = e[i].to;
    if (pre[x] == to) continue;
    pre[to] = x;
    deep[to] = deep[x] + 1;
    dfs1(to);
    size[x] += size[to];
    if (size[to] > size[son[x]]) son[x] = to;
  }
}
void dfs2(int x, int topx) {
  pos[x] = ++tree_idx, top[x] = topx;
  if (son[x]) dfs2(son[x], topx);
  for (int i = head[x]; i != -1; i = e[i].next)
    if (e[i].to != son[x] && e[i].to != pre[x]) dfs2(e[i].to, e[i].to);
}
node pd_l(node x, node y, int lim) {
  node p;
  int ll = x.num, rr = y.num, ld, rd, val;
  ld = rd = 1;
  while (p.num < lim && (ld <= ll || rd <= rr)) {
    if (ld > ll)
      val = y.ps[rd], rd++;
    else if (rd > rr)
      val = x.ps[ld], ld++;
    else {
      if (x.ps[ld] < y.ps[rd])
        val = x.ps[ld], ld++;
      else
        val = y.ps[rd], rd++;
    }
    p.ps[++p.num] = val;
  }
  return p;
}
void build(int lr, int l, int r) {
  if (l == r) {
    ans[lr] = dp[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(lr << 1, l, mid);
  build(lr << 1 | 1, mid + 1, r);
  ans[lr] = pd_l(ans[lr << 1], ans[lr << 1 | 1], 10);
}
node query(int ll, int rr, int v, int lr, int l, int r) {
  if (ll <= l && r <= rr) {
    return ans[lr];
  }
  int mid = (l + r) >> 1;
  if (rr <= mid)
    return query(ll, rr, v, lr << 1, l, mid);
  else if (ll > mid)
    return query(ll, rr, v, lr << 1 | 1, mid + 1, r);
  else {
    node x, y, z;
    x = query(ll, rr, v, lr << 1, l, mid);
    y = query(ll, rr, v, lr << 1 | 1, mid + 1, r);
    z = pd_l(x, y, v);
    return z;
  }
}
node all_query(int x, int y, int z) {
  node p, vs;
  while (top[x] != top[y]) {
    if (deep[top[x]] < deep[top[y]]) swap(x, y);
    int a = top[x];
    vs = query(pos[a], pos[x], z, 1, 1, n);
    p = pd_l(p, vs, z);
    x = pre[a];
  }
  if (deep[x] > deep[y]) swap(x, y);
  vs = query(pos[x], pos[y], z, 1, 1, n);
  p = pd_l(p, vs, z);
  return p;
}
int main() {
  init();
  int x, y, z;
  node p;
  n = read(), m = read(), q = read();
  for (int i = 1; i < n; i++) {
    x = read(), y = read();
    addedges(x, y);
    addedges(y, x);
  }
  deep[1] = 0;
  dfs1(1);
  dfs2(1, 1);
  for (int i = 1; i <= m; i++) {
    x = read();
    if (dp[pos[x]].num < 10) dp[pos[x]].ps[++dp[pos[x]].num] = i;
  }
  build(1, 1, n);
  while (q--) {
    x = read(), y = read(), z = read();
    p = all_query(x, y, z);
    printf("%d", p.num);
    for (int i = 1; i <= p.num; i++) printf(" %d", p.ps[i]);
    puts("");
  }
  return 0;
}